Diophantine equations

[paradox]

i was looking at a TMSCA test and wondered upon a diophantine equation
given all the cases such as x and y are both positive or negative or either is negative or positive
how do you solve these thanks

[FabensMath]

I was told to take the least common multiple of the x and y values, and divide the other number by this, then subtract 1 to get the answer.

[007math]

[chrishuff]

I'm guessing that you would round down in that case...

[rcoker7]

There's different ways to solve each different case to this problem.

case 1: the problem says you need to find the number of POSITIVE integers.

Ok let's say we have ax+by=c. If neither a or b divide c, then you don't always do the same thing. The exact answer for this is c/(ab)+1-(min(x)/b+min(y)/a) where min(x) is the smallest value of x that works and min(y) is the smallest value of y that works. So basically you have c/(ab) either rounded up or down, depending on the numbers.

If one of a or b divides c, then you take c/(ab) and round it down. If both divide c, then you round it down then subtract 1.

case 2: the problem says you need to find the number of NONNEGATIVE integers.

This works the exact same when neither a or b divide c.

If either a or b(or both) divide c then you simply take c/(ab) and round up.

sorry for the 3 day bump and the lack of latex, btw.

EDIT: This only works if a and b are relatively prime.

[FabensMath]

Thanks Ross! Your'e awesome!